Question: Solve for $r$, $ \dfrac{6}{8r} = \dfrac{5r + 8}{20r} - \dfrac{5}{4r} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $8r$ $20r$ and $4r$ The common denominator is $40r$ To get $40r$ in the denominator of the first term, multiply it by $\frac{5}{5}$ $ \dfrac{6}{8r} \times \dfrac{5}{5} = \dfrac{30}{40r} $ To get $40r$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ \dfrac{5r + 8}{20r} \times \dfrac{2}{2} = \dfrac{10r + 16}{40r} $ To get $40r$ in the denominator of the third term, multiply it by $\frac{10}{10}$ $ -\dfrac{5}{4r} \times \dfrac{10}{10} = -\dfrac{50}{40r} $ This give us: $ \dfrac{30}{40r} = \dfrac{10r + 16}{40r} - \dfrac{50}{40r} $ If we multiply both sides of the equation by $40r$ , we get: $ 30 = 10r + 16 - 50$ $ 30 = 10r - 34$ $ 64 = 10r $ $ r = \dfrac{32}{5}$